Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
    private static double solve(int N, double[] p) {
        double[] hCount = new double[N+1];
        hCount[0] = 1.0;
        for (int i = 0; i < N; i++)
            for (int j = i+1; j >=0; j--)
                hCount[j] = ((j >= 1) ? p[i]*hCount[j-1]: 0) + (1.0-p[i])*hCount[j];
        double sum = 0.0;
        for (int i = N/2+1; i <= N; i++) sum += hCount[i];
        return sum;
    }

    public static double run(Scanner scanner) {
        int N = scanner.nextInt();
        double[] p = new double[N];
        for (int i=0; i < N; i++) p[i] = scanner.nextDouble();

        return solve(N, p);
    }

    public static void main(String[] args) {
        try (Scanner scanner = new Scanner(System.in)) {
            System.out.println(run(scanner));
        }
        //Tests.run();
    }
}